# Difference between revisions of "1969 Canadian MO Problems/Problem 3"

(→Solution: SOLVE INEQUALITIES IN THE PROPER DIRECTION!!!!!!!) |
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== Solution == | == Solution == | ||

− | + | By the [[Pythagorean Theorem]] and the [[trivial inequality]], <math>\displaystyle 2c^2-(a+b)^2=2(a^2+b^2)-(a+b)^2=(a-b)^2\ge 0</math>. | |

− | + | Thus <math>\displaystyle 2c^2\ge (a+b)^2.</math> Since <math>\displaystyle a,b,c</math> are all positive, taking a square root preserves the inequality and we have our result. | |

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+ | The equality condition is clearly that <math>(a-b)^2 = 0</math> -- the [[isosceles]] [[right triangle]]. | ||

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## Revision as of 09:21, 28 July 2006

## Problem

Let be the length of the hypotenuse of a right angle triangle whose two other sides have lengths and . Prove that . When does the equality hold?

## Solution

By the Pythagorean Theorem and the trivial inequality, .

Thus Since are all positive, taking a square root preserves the inequality and we have our result.

The equality condition is clearly that -- the isosceles right triangle.